Find the area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

$\pi ab$

The correct answer is Option (3) → $\pi ab$

From given figure, the area of the region $ABA'B'A$ bounded by the ellipse

$= 4 \left( \text{area of the region AOBA in the first quadrant bounded by the curve, } x\text{-axis and the ordinates } x = 0, x = a \right)$

(as the ellipse is symmetrical about both $x$-axis and $y$-axis)

$= 4 \int\limits_{0}^{a} y \, dx \quad \text{(taking vertical strips)} \text{}$

Now $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ gives $y = \pm \frac{b}{a} \sqrt{a^2 - x^2}$, but as the region AOBA lies in the first quadrant, $y$ is taken as positive. So, the required area is

$= 4 \int\limits_{0}^{a} \frac{b}{a} \sqrt{a^2 - x^2} \, dx \text{}$

$= \frac{4b}{a} \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1} \frac{x}{a} \right]_{0}^{a} $

$= \frac{4b}{a} \left[ \left( \frac{a}{2} \times 0 + \frac{a^2}{2}\sin^{-1} 1 \right) - 0 \right] \text{}$

$= \frac{4b}{a} \frac{a^2}{2} \frac{\pi}{2} = \pi ab \text{}$